Question: Let $y=\dfrac1x\sin(x)$. $\dfrac{dy}{dx}=$
Explanation: $\dfrac1x\sin(x)$ is the product of two, more basic, expressions: $\dfrac1x$ and $\sin(x)$. Therefore, $\dfrac{dy}{dx}$ can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\dfrac1x\sin(x)\right) \\\\ &=\dfrac{d}{dx}\left(\dfrac1x\right)\sin(x)+\dfrac1x\dfrac{d}{dx}(\sin(x))&&\gray{\text{The product rule}} \\\\ &=\dfrac{d}{dx}\left(x^{-1}\right)\sin(x)+\dfrac1x\dfrac{d}{dx}(\sin(x))&&\gray{\text{Write } \dfrac1x\text{as a power}} \\\\ &=-1x^{-2}\cdot \sin(x)+\dfrac1x\cdot \cos(x)&&\gray{\text{Differentiate }x^{-1}\text{ and }\sin(x)} \\\\ &=\dfrac{\cos(x)}{x}-\dfrac{\sin(x)}{x^2}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{\cos(x)}{x}-\dfrac{\sin(x)}{x^2}$ or any other equivalent form.